HDU - 5372 Segment Game 树状数组

HDU - 5372

我们考虑总的线段减去违法的线段， 因为是按长度从小到大加入数轴的， 所以违法线段就是Rj > Ri的个数加上 Lj < Li 的个数， 

用树状数组维护就完事了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 4e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n;
int addtot;
int X[N], Xtot;
int ans[N];
int id[N];
int L[N], R[N];
int op[N], b[N];

struct Bit {
    int a[N];
    void init() {
        for(int i = 1; i <= Xtot; i++) {
            a[i] = 0;
        }
    }
    inline void modify(int x, int v) {
        for(int i = x; i <= Xtot; i += i & -i) {
            a[i] += v;
        }
    }
    inline int sum(int x) {
        int ans = 0;
        for(int i = x; i; i -= i & -i) {
            ans += a[i];
        }
        return ans;
    }
    inline int query(int L, int R) {
        if(L > R) return 0;
        return sum(R) - sum(L - 1);
    }
} bit[2];


void init() {
    Xtot = addtot = 0;
}

int main() {

    int cas = 0;

    while(scanf("%d", &n) != EOF) {
        init();
        int now = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d%d", &op[i], &b[i]);
            if(!op[i]) {
                addtot++;
                ans[addtot] = now;
                id[i] = addtot;
                L[addtot] = b[i];
                R[addtot] = b[i] + addtot;
                X[++Xtot] = b[i];
                X[++Xtot] = b[i] + addtot;
                now++;
            }
            else {
                now--;
            }
        }
        sort(X + 1, X + 1 + Xtot);
        Xtot = unique(X + 1, X + 1 + Xtot) - X - 1;
        for(int i = 1; i <= addtot; i++) {
            L[i] = lower_bound(X + 1, X + 1 + Xtot, L[i]) - X;
            R[i] = lower_bound(X + 1, X + 1 + Xtot, R[i]) - X;
        }

        bit[0].init(); bit[1].init();

        for(int i = 1; i <= n; i++) {
            if(op[i] == 0) {
                ans[id[i]] -= bit[0].query(1, L[id[i]] - 1);
                ans[id[i]] -= bit[1].query(R[id[i]] + 1, Xtot);
                bit[0].modify(L[id[i]], 1);
                bit[1].modify(R[id[i]], 1);
            }
            else {
                bit[0].modify(L[b[i]], -1);
                bit[1].modify(R[b[i]], -1);
            }
        }

        printf("Case #%d:\n", ++cas);
        for(int i = 1; i <= addtot; i++) {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}

/*
*/